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If fourth term of an A.P. is zero, then \(\rm \dfrac{t_{25}}{t_{11}}\) is, where tn denotes the nth term of AP.
1. 2
2. 3
3. 4
4. 5
5. None of these

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Correct Answer - Option 2 : 3

Concept:

Let us consider sequence a1, a2, a3 …. an is an A.P.

  • Common difference “d”= a2 – a1 = a3 – a2 = …. = an – an – 1
  • nth term of the A.P. is given by an = a + (n – 1) d
  • Sum of the first n terms = Sn =\(\rm \frac n 2\) [2a + (n − 1) × d]= \(\rm \frac n 2\)(a + l)


Where, a = First term, d = Common difference, n = number of terms, an = nth term and l = Last term

Calculation:

Let the first term of AP be 'a' and the common difference be 'd'

Given: Fourth term of an A.P. is zero

⇒ a4 = 0

⇒ a + (4 - 1) × d = 0

⇒ a + 3d = 0         

∴ a = -3d                     .... (1)

To Find: \(\rm \dfrac{t_{25}}{t_{11}}\)

\(\rm \Rightarrow \dfrac{t_{25}}{t_{11}} = \dfrac {a+24d}{a+10d}\\=\dfrac {-3d+24d}{-3d+10d}\\=\dfrac {21d} {7d}=3\)

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